the difference of two figures is 8. thediff of two squares is 400. tell what it is?

Answer 1

these two figures are 21 and 29.there difference is 8.and there squares are 441 and 841 respectively ,and the difference of squares is 400.

Answer 2

The equation can be solved easily, as long as you remember to substitute at the right time.

We know:

x^2 - y^2 =400
x - y = 8

x^2 - y^2 = 400 Factor the squared terms
(x + y) ( x - y) = 400
We know from the first statements that x-y = 8, so substitute the 8 into the above equation:
8( x + y) = 400 Divide by both side by 8
x + y = 50 Add this equation to the one we know:
x + y = 50
+x - y = 8
--------------------
2x = 58 so x = 29 Now plug the value of x into the second equation we know:
x - y = 8 29 - 8 = y therefore , y = 21

x = 29 and y = 21
Also, you can plug both values into the second equation, and you will see it works.

Answer 3

The best way to answer this question is to use algebra. Call the two numbers "a" and "b":

a-b=8

a^2 - b^2 = 400

You can solve these unknowns by the following method:

Rearrange the first equation to solve for "a":
a=8+b

Substitute into the second equation:

(8+b)^2 - b^2=400

Expand:
64 + 16b +b^2 -b^2 =400

Simplify:
64 +16b =400

Subtract 64 from each side:
16b=336

Divide both sides by 16:
b=21

Substitute into the modified first equation:
a=8+21

a= 29

Check the equations to make sure:
29 - 21= 8 good
29^2 - 21^2 = 400 good

Ta da!! a=29 and b=21

Answer 4

x-y=8
x=y+8

x^2-y^2=400
(y+8)^2-y^2=400
(y+8)(y+8)-y^2=400
y^2+8y+8y+64-y^2=400
y^2-y^2=0 SO
8y+8y+64=400
16y=336
y=21

Substitue y into original equation.
x-21=8
x=29

The two figures are 21 and 29. Their difference is 8.
29-21=8. The difference of the squares is 400.
841-441=400.

Answer 5

29 and 21

Answer 6

21 and 29

Answer 7

Let the figures be x and y.
We have the equations
x-y=8
X^2-y^2=400
now it is very simple
x^2-y^2=(x+y)(x-y)=400
therefore x+y=400/8=50
y=50-x
x=y+8=50-x+8
2x=58
x=29
y=21
verification
x-y=8 ok
x^2-Y^2=841-441=400 ok

Answer 8

Let the two numbers be a and b.
As per given condition:
a-b=8..............(1) and a^2-b^2=400...................(2)
a^2-b^2=400
(a+b)*(a-b)=400
(a+b)*8=400 .........................from equation (1)
.:.(a+b)=400/8=50 .................equation(3)
from equation(1) and (3)
a+b=50
a-b=8
----------
2a=58
.:.a=29
after substituting a=29 in equation(3) we get value of b as 11.
Hence these numbers are 29 and 11

Answer 9

x - y = 8
x^2 - y^2 = 400

let x = 8 + y

replacing x
(8 + y)^2 - y^2 = 400
expanding bracket
64 + 16y +y^2 - y^2 = 400
64 + 16y = 400
16y = 336
y= 21

replacing y in x = 8 + y
x = 29

Answer 10

The differende of two figures is 8

x - y = 8

The difference of two squares if 400

x² - y² = 400